Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $q = \dfrac{10a - 10}{a + 5} \div \dfrac{a^2 - 3a + 2}{a^2 + 5a} $
Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{10a - 10}{a + 5} \times \dfrac{a^2 + 5a}{a^2 - 3a + 2} $ First factor the quadratic. $q = \dfrac{10a - 10}{a + 5} \times \dfrac{a^2 + 5a}{(a - 1)(a - 2)} $ Then factor out any other terms. $q = \dfrac{10(a - 1)}{a + 5} \times \dfrac{a(a + 5)}{(a - 1)(a - 2)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ 10(a - 1) \times a(a + 5) } { (a + 5) \times (a - 1)(a - 2) } $ $q = \dfrac{ 10a(a - 1)(a + 5)}{ (a + 5)(a - 1)(a - 2)} $ Notice that $(a + 5)$ and $(a - 1)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ 10a\cancel{(a - 1)}(a + 5)}{ (a + 5)\cancel{(a - 1)}(a - 2)} $ We are dividing by $a - 1$ , so $a - 1 \neq 0$ Therefore, $a \neq 1$ $q = \dfrac{ 10a\cancel{(a - 1)}\cancel{(a + 5)}}{ \cancel{(a + 5)}\cancel{(a - 1)}(a - 2)} $ We are dividing by $a + 5$ , so $a + 5 \neq 0$ Therefore, $a \neq -5$ $q = \dfrac{10a}{a - 2} ; \space a \neq 1 ; \space a \neq -5 $